The trapezium rule Problem set B One question at a time

Use the trapezium rule with 6 strips to estimate \(\int_{0}^{3} \frac{1}{x+6}\,dx\).

• \(h=\dfrac{3-0}{6}=0.5\)
• x: 0, 0.5, 1, 1.5, 2, 2.5, 3; y: 0.1667, 0.1538, 0.1429, 0.1333, 0.1250, 0.1176, 0.1111
• \(\tfrac12(0.5)[(0.1667+0.1111)+2(0.1538+0.1429+0.1333+0.1250+0.1176)]\) = 0.4058

Use the trapezium rule with 6 strips to estimate \(\int_{2}^{5} \frac{1}{x}\,dx\).

• \(h=\dfrac{5-2}{6}=0.5\)
• x: 2, 2.5, 3, 3.5, 4, 4.5, 5; y: 0.5000, 0.4000, 0.3333, 0.2857, 0.2500, 0.2222, 0.2000
• \(\tfrac12(0.5)[(0.5000+0.2000)+2(0.4000+0.3333+0.2857+0.2500+0.2222)]\) = 0.9206

Use the trapezium rule with 5 strips to estimate \(\int_{0}^{1.5} 5e^x\,dx\).

• \(h=\dfrac{1.5-0}{5}=0.3\)
• x: 0, 0.3, 0.6, 0.9, 1.2, 1.5; y: 5.000, 6.749, 9.111, 12.30, 16.60, 22.41
• \(\tfrac12(0.3)[(5.000+22.41)+2(6.749+9.111+12.30+16.60)]\) = 17.54

Use the trapezium rule with 6 strips to estimate \(\int_{4}^{7} \ln x\,dx\).

• \(h=\dfrac{7-4}{6}=0.5\)
• x: 4, 4.5, 5, 5.5, 6, 6.5, 7; y: 1.386, 1.504, 1.609, 1.705, 1.792, 1.872, 1.946
• \(\tfrac12(0.5)[(1.386+1.946)+2(1.504+1.609+1.705+1.792+1.872)]\) = 5.074

Use the trapezium rule with 4 strips to estimate \(\int_{1}^{3} \ln(6x)\,dx\).

• \(h=\dfrac{3-1}{4}=0.5\)
• x: 1, 1.5, 2, 2.5, 3; y: 1.792, 2.197, 2.485, 2.708, 2.890
• \(\tfrac12(0.5)[(1.792+2.890)+2(2.197+2.485+2.708)]\) = 4.866

The table shows the speed v (m/s) of a cyclist, measured every 10 seconds. The area under the graph is the distance travelled. t: 10, 20, 30, 40, 50, 60 v: 3.2, 5.1, 6.4, 6.9, 6, 4.4 Estimate the area under the graph.

• \(h=10\)
• \(\tfrac12(10)[(3.200+4.400)+2(5.100+6.400+6.900+6.000)]\) = 282.0
• Answer ≈ 282

Water flows into a tank at a rate R (litres per minute), measured every 5 minutes. t: 0, 5, 10, 15, 20 R: 12, 9.4, 7.9, 6.1, 5.2 Estimate the volume of water that flows in during the 20 minutes.

• \(h=5\)
• \(\tfrac12(5)[(12.00+5.200)+2(9.400+7.900+6.100)]\) = 160.0
• Answer ≈ 160 litres

Oil leaks from a pipe at the rate \(r=\ln(5t)\) litres per hour, for \(1\le t\le 3.5\), where t is in hours. Use the trapezium rule with 5 strips to estimate the volume that leaks between t = 1 and t = 3.5.

• \(h=0.5\)
• x: 1, 1.5, 2, 2.5, 3, 3.5; y: 1.609, 2.015, 2.303, 2.526, 2.708, 2.862
• \(\tfrac12(0.5)[(1.609+2.862)+2(2.015+2.303+2.526+2.708)]\) = 5.894

The integral \(\int_{2}^{7} \frac{1}{x}\,dx\) is to be estimated.

1. (a) Use the trapezium rule with 5 strips.

   • \(h=\dfrac{7-2}{5}=1\)
   • x: 2, 3, 4, 5, 6, 7; y: 0.5000, 0.3333, 0.2500, 0.2000, 0.1667, 0.1429
   • \(\tfrac12(1)[(0.5000+0.1429)+2(0.3333+0.2500+0.2000+0.1667)]\) = 1.271

2. (b) Use the trapezium rule with 10 strips.

   • \(h=\dfrac{7-2}{10}=0.5\)
   • x: 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7; y: 0.5000, 0.4000, 0.3333, 0.2857, 0.2500, 0.2222, 0.2000, 0.1818, 0.1667, 0.1538, 0.1429
   • \(\tfrac12(0.5)[(0.5000+0.1429)+2(0.4000+0.3333+0.2857+0.2500+0.2222+0.2000+0.1818+0.1667+0.1538)]\) = 1.258

3. (c) Use your calculator to find the value of the integral. Which estimate is closer to it, and why?

   • Calculator: \(\int_{2}^{7}\frac1x\,dx = 1.253\)
   • Differences: 0.0187 for 5 strips, 0.00475 for 10 strips
   • More strips fit the curve more closely.

The integral \(\int_{0}^{3} 2e^x\,dx\) is to be estimated.

1. (a) Use the trapezium rule with 3 strips.

   • \(h=\dfrac{3-0}{3}=1\)
   • x: 0, 1, 2, 3; y: 2.000, 5.437, 14.78, 40.17
   • \(\tfrac12(1)[(2.000+40.17)+2(5.437+14.78)]\) = 41.30

2. (b) Use the trapezium rule with 6 strips.

   • \(h=\dfrac{3-0}{6}=0.5\)
   • x: 0, 0.5, 1, 1.5, 2, 2.5, 3; y: 2.000, 3.297, 5.437, 8.963, 14.78, 24.36, 40.17
   • \(\tfrac12(0.5)[(2.000+40.17)+2(3.297+5.437+8.963+14.78+24.36)]\) = 38.96

3. (c) Find the value of the integral with your calculator. By how much does each estimate differ from it?

   • Calculator: \(\int_{0}^{3}2e^x\,dx = 38.17\)
   • Differences: 3.13 and 0.792